Is ${166839}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {166839}= &&{1}\cdot100000+ \\&&{6}\cdot10000+ \\&&{6}\cdot1000+ \\&&{8}\cdot100+ \\&&{3}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {166839}= &&{1}(99999+1)+ \\&&{6}(9999+1)+ \\&&{6}(999+1)+ \\&&{8}(99+1)+ \\&&{3}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {166839}= &&\gray{1\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {1}+{6}+{6}+{8}+{3}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${166839}$ is divisible by $3$ if ${ 1}+{6}+{6}+{8}+{3}+{9}$ is divisible by $3$ Add the digits of ${166839}$ $ {1}+{6}+{6}+{8}+{3}+{9} = {33} $ If ${33}$ is divisible by $3$ , then ${166839}$ must also be divisible by $3$ ${33}$ is divisible by $3$, therefore ${166839}$ must also be divisible by $3$.